TITLE: Is there any deep reason that 23456789 is prime?
QUESTION [7 upvotes]: I was recently coding a prime factorization function and wanted to test that it could factor large numbers reasonably well. Arbitrarily, I slid my finger across the number pad, and got 23456789 as a test input. Surprisingly, it was prime!
Was this just a strange coincidence, or is there any deeper number theoretic structure that leads to this result.
Past research: all I've been able to do is confirm that 23456789 is indeed prime and is the largest prime with digits in ascending order. https://primes.utm.edu/curios/page.php/23456789.html

REPLY [2 votes]: COMMENT.-Here something I like and and maybe related to the problem proposed (note that with this definition the coefficient $a$ can be extended to be non-digit).
Define
$$P_n(x)=\sum_{2\le a\le n} ax^{n-a}$$ 
The polynomials $P_n(x)$ have degree $n-2$ and can be defined by recurrence via
$$P_{n+1}(x)=xP_n(x)+(n+1);\space P_2(x)=2$$ This way one has
$$P_9(10)=\color{red}{23456789}\\P_9(4)=50969,\text{ prime }\\P_9(2)=757,\text{ prime}$$ Besides with exception of $P_6$ for which $P_6(n)$ is composite for $1\le n\le 10$ one has the following primes
for $1\le n\le 10$
$$\begin{cases}P_8(5)=43943\\P_8(3)=2729\\P_7(3)=907\\P_5(2)=41\\P_5(6)=569\\P_4(3)=31\\P_4(9)=193\\P_3(1)=5\\P_3(2)=7\\P_3(4)=11\\P_3(5)=13\\P_3(7)=17\\P_3(8)=19\\P_3(10)=23\\P_2(n)=2\text{ for all } n\end{cases}$$